Cosmology Cosmological Nucleosynthesis

O In the discussion of the thermal history of the early universe form temperatures 1011 ◦K down to about 106 ◦K we were neglecting the tiny contribution of relict nucleons which were not completely annihilated in particle-antiparticle annihilation.

O Nucleons, the building blocks of atomic nuclei, were formed abundantly in the QCD phase transition out of the quark-gluon plasma at temperatures T ∼ 150 MeV corresponding to a temperature T ∼ 2 × 1012 ◦K and cosmic time t ∼ 20µs.

O The QCD phase transition sets in after the annihilation of the heavier quarks (t, b, c) during the annihilation of s quarks. During the QCD phase transition color neutral SU(3)c triplet states form, the baryons, as well as anti-triplets representing the antibaryons together with quark–antiquark singlets, the mesons. The lightest hadrons are the nucleons (N), proton (p) and neutron (n), and their antiparticles (spin 1/2 fermions), which when formed are non-relativistic (essentially at rest) and the lightest mesons, the pions π± and π0 (spin 0 bosons).

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 138 Cosmology

O Baryon–antibaryon annihilation immediately sets in during their formation. Here the mystery of the baryon asymmetry plays the key role for the future evolution, namely, the annihilation turned out to be incomplete with a relict −10 −9 η = nN(t0)/nγ(t0) ∼ 10 − 10 of surviving baryons. The only particles left in large numbers at this time were pions, muons, electrons, neutrinos and photons.

After baryon–antibaryon annihilation, for T < 10 MeV, the total number of baryons nB = nN − nN¯ = np − np¯ + nn − nn¯, in practice nB = np + nn since now nN¯  nN, remains conserved due to baryon number conservation. Baryons are free protons, free neutrons or protons and neutrons bound in nuclei. While nuclear strong interaction forces bind nucleons in nuclei weak interactions convert neutron and protons into each other.

Ì As we will see, about one quarter of all nucleons form nuclei (A > 1), with about an equal amount of neutrons and protons, while approximately three quarters remain free protons.

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 139 Cosmology

+ − Ì Later, when T < 10 keV, after e e –annihilation, we also will have ne+  ne− and ne− = np, where np represents all protons, the free ones as well as the ones bound in nuclei (presumed charge neutrality of the universe).

Ì Radiation and matter (electron, protons and nuclei) remain in thermal equilibrium as long as there are lots of free electrons. After e+e−–annihilation, atoms in particular hydrogen may form

e− + p H + γ and essentially, when Tγ < 13.6 eV, the ionization energy of hydrogen, electrons will have been trapped by protons and nuclei ions (recombination). The universe becomes transparent and photons decouple from matter. More details on that later.

We are now discussing the evolution of the relict densities of baryons and the formation of light nuclei.

Relevant basic processes:

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 140 Cosmology

n + ν p + e− , n + e+ p + ν¯ , n p + e− + ν¯

O ν ≡ νe, other neutrinos inactive here.

O kB T  mN → nucleons at rest (non-relativistic)

O Q–value: Q = mn − mp ' 1.293 MeV

− Q = Ee − Eν n + ν p + e + Q = Eν − Ee n + e p + ν¯ − Q = Eν + Ee n p + e + ν¯

Calculation of rate:

Ì cross-section: weak 2–body reaction n + e → p + ν¯

2 2 2 2π2~3 AE2 G (1+3 g ) cos θC σ = ν ; A = F A ve 2π3 ~

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 141 Cosmology

−5 −2 GF = 1.16637(1) × 10 GeV Fermi constant gA = 1.257 nucleon axial vector coupling of β-decay cos θC = 0.9745(6) , ΘC Cabibbo mixing angle ve the electron velocity

Electrons and neutrinos are relativistic → have to take into account Fermi distribution (Pauli-Principle):

Ì number density of electrons (with momentum between pe and pe + dpe)

2 4 π pe dpe exp (E /k T) + 1 −1 (2 π ~)3 e B

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 142 Cosmology

Ì fraction of unfilled antineutrino levels:

 −1  −1 1 − exp (Eν/kBTν) + 1 = exp (−Eν/kBTν) + 1 such that the total rate reads

∞ + R 2 2  −1  −1 λ(n + e → p + ν¯) = A Eν pe dpe exp (Ee/kBT) + 1 × exp (−Eν/kBTν) + 1 0 changing variable to q ≡ −Eν = −Q − Ee yields an integral over q

+ O from q = −∞ to q = −Q − me for n + e → p + ν¯ − O from q = −Q + me to q = 0 for n → p + e + ν¯ O from q = 0 to q = +∞ for n + ν → p + e− while the integrand is unchanged.

The total rate of disappearance of neutrons then is given by

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 143 Cosmology

+∞ r R m2 (Q+q)2 q2 dq λ(n → p) = = 1 − e     (Q+q)2 1+eq/kBTν 1+e−(Q+q)/kBT −∞ with a gap from q = −Q − me to q = −Q + me in the integration range (unphysical region).

Similarly, one calculates the total rate of production of neutrons to be

+∞ r R m2 (Q+q)2 q2 dq λ(p → n) = = 1 − e     (Q+q)2 1+e−q/kBTν 1+e(Q+q)/kBT −∞ again, with a gap from q = −Q − me to q = −Q + me in the integration range.

This allows us to calculate the fraction of neutrons among the nucleons, by solving the differential equation

dXn dt = −λ(n → p) Xn + λ(p → n) (1 − Xn)

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 144 Cosmology with appropriate initial conditions.

Initial Condition

One can check (Exercise) that for fixed time–independent temperature T = Tν the two rates λ(p → n) and λ(n → p) have a ratio λ(p→n) λ(n→p) = exp (−Q/kBT) for T = Tν which would imply a time–independent solution

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 145 Cosmology Xn = Xn = exp (−Q/k T) , Xp 1−Xn B the non-relativistic equilibrium distribution. Thus what keeps this ratio away from equilibrium is the inequality of T and Tν and the fact that they depend on time. When baryons are produced during the QCD–phase transition we have kBT  Q and we can calculate the integrals by setting T = Tν and Q, me = 0, such that

+∞ Z q4 dq λ(n → p) = λ(p → n) = A

1 + eq/kBT 1 + e−q/kBT −∞ 7  T 5 = π4 A (k T)5 = 0.400 sec−1 15 B 1010 ◦K which is > 1 for T > 1.22 × 1010 ◦K .

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 146 Cosmology

Thermal equilibrium means dXn λ(p→n) 1 = 0 ⇔ Xn = ⇔ Xn = dt λ(p→n)+λ(n→p) (1+exp(Q/kBT))

Crucially important is that the equilibrium value is established at high enough temperatures when the electron chemical potential is zero (still large numbers of electrons and positrons) such that no assumption about the initial proton/neutron ratio is necessary!

This is expected to be satisfied for T > 3 × 1010 ◦K and we may solve our exact equation for Xn with this boundary condition. The solution can be obtained numerically; see the Table

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 147 Cosmology

◦ T( K) t(sec) Xn 1 × 1012 0.0001 0.4962 3 × 1011 0.0011 0.4875 1 × 1011 0.0099 0.4626 3 × 1010 0.1106 0.3798 1 × 1010 1.008 0.2386 3 × 109 12.67 0.1654 1.3 × 109 91.09 0.1458 1.2 × 109 110.2 0.1425 1.1 × 109 135.1 0.1385 1 × 109 168.1 0.1333 9 × 108 2.12.7 0.1268 8 × 108 274.3 0.1182 7 × 108 362.6 0.1070 6 × 108 496.3 0.0919 3 × 108 1980 0.0172 1 × 108 17780 3.07 × 10−10

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 148 Cosmology

Given the rate λ, the maximum number of collisions can be found by multiplying the rate with the Hubble time H = 1/2t. Earlier, we calculated

s !2 3 1 1010 ◦K t = + const = 0.994 sec + const 16πGag∗ T 2 T

 3 such that, with g∗ = 43/4, λ/H ' 0.8 × T which is > 1 for 1010 ◦K T > 1.1 × 1010 ◦K . Because of the T 3 behavior the rate drops fast and equilibrium is lost quickly between 3 × 1010 ◦K and 1010 ◦K . Little later all but neutron β–decay plays a role in proton– neutron transmutation, also due to the fact that electron and positron densities rapidly decrease because of e+e−–annihilation. Then the decrease of Xn just follows neutron decay.

Exercise: Show that Xn → 0.1609 exp (−t/τn) with τn = 885.7 sec is the neutron lifetime. The above would be the story if nucleons would not get bound into nuclei where also neutrons get stable.

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 149 Cosmology

Nuclei if in Equilibrium

We are interested here in what happens after temperatures have dropped way below the proton mass T  mp such that all nucleons and nuclei are non-relativistic. In thermal equilibrium we then have (see Exercise at end of previous lecture) m T 3/2 µi−mi i T ni = gi 2π e for the number densities of proton (i = p), neutron (i = n) or a nuclei of type i = (Ai − Zi) n + Zi p.

If nuclear reaction rates for the formation of a nucleus (Ai − Zi) n + Zi p i from protons and neutrons would be sufficiently high to maintain chemical equilibrium, we would have

µi = (Ai − Zi) µn + Zi µp

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 150 Cosmology for the chemical potentials. Since for free nucleons gi = 2 for i = p, n we have

3 3(A −1) 2 −A  2π 2 i Zi Ai−Zi B /T ni = gi A 2 i n n e i i mNT p n

where

Bi = Zi mp + (Ai − Zi) mn − mi

is the binding energy of the nucleus. Besides in exponents, we use the nucleon mass mN as an approximation mp = mn = mi/Ai = mN.

Nuclear binding energies of light nuclei

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 151 Cosmology

A 2 iZi Bi gi Ti for ΩBh ' 0.2 1H 13.6 eV (atomic binding energy) 2H 2.22 MeV 3 0.75 × 109 ◦K 3H 8.48 MeV 2 1.4 × 109 ◦K 3He 7.72 MeV 2 1.3 × 109 ◦K 4He 28.3 MeV 1 3.3 × 109 ◦K 7Li 39.24 MeV 4 2.3 × 109 ◦K 7Be 37.60 MeV 4 2.2 × 109 ◦K 12C 92.2 MeV 1 2.9 × 109 ◦K

Total baryon number density now is: P nB = i ni Its present value is usually encoded in the ratio NB(T0) 2 ζ(3) 3 ◦ η = ; nγ(T) = T ; T0 = 2.725 K nγ(T0) π2

Baryon number conservation and the scaling property n(t) ∝ S (t)−3 allow us to

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 152 Cosmology calculate nB at temperature T. We may also use the fact that the entropy is 3 1 2π2 3 constant S (t) s(t) = const with s(t) = T(t) (ρ(t) + p(t)) = g∗s(T) 45 T (t) or 3 3 g∗s(T) S (t) T (t) = const: 2ζ(3) g∗s(T) 3 nB = η T π2 g∗s(T0) as long as baryon number is conserved and the universe expands adiabatically. For T  m we have the simpler relation n η n η 2ζ(3) T 3. e B = γ = π2 In order to get ride of the unknown chemical potentials one considers ni Zi Ai−Zi np nn and represents the equilibrium distributions in terms of dimensionless ratios

Xi ≡ ni/nN , Xp ≡ np/nN , Xn ≡ nn/nN where nN is the number density of all nucleons. Then

3 gi Zi Ai−Zi 2 (Ai−1) Bi/T Xi = 2 Xp Xn Ai ε e

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 153 Cosmology where

 −3 3 S (t)  −3/2 1 3 −2 −11 T 2 ε ≡ nNh (2πmNT) = 2.96 × 10 10 10 ◦ ΩB h 2 10 S (t0) 10 K

2 3 with nN = 3ΩB H0 (S/S 0) /8πG mN the total number density of nucleons.

In the period of interest after e+e−–annihilation T ∝ 1/S such that  +3/2 ε = 1.46 × 10−12 T Ω h2 1010 ◦K B

Thus ε is very small such that in equilibrium the nuclear species i are essentially absent until the binding term takes over for sufficiently small temperatures: Bi Ti ' . (Ai−1) | ln ε|

2 where the Ti are given in the Table above. The dependence on ΩBh is weak. 4 Heavier elements have Ti’s similar to He as they have similar binding energy per nucleon.

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 154 Cosmology

If thermal and chemical equilibrium would have been maintained from temperatures of about 3 × 1010 ◦K down to 109 ◦K then during this time He4 and the heavier elements would have been built first followed by the lighter ones He3

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 155 Cosmology and H3, which would decay later into He3 and finally H2. This is not what has happened! The reason: densities were too low for any other than two–body reactions to take place!

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 156 Cosmology

The Chain of 2–body processes

1) p + n → d + γ 2) d + d → H3 + p , He3 + n 3) d + H3 → He4 + n ; d + He3 → He4 + p 4) slower processes involving γ’s

O Step 1) no problem, rate of deuterium production per free neutron

!−3 S (t) λ . × −20 3/ × n −1 X h2 d = 4 55 10 cm sec p = 511 sec −9 pΩB 10 S 0  T  = 2.52 × 104 sec−1 X Ω h2 1010 ◦K p B

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 157 Cosmology

In this period, after e+e−–annihilation we have g∗ = 3.363 and the time vs. temperature relation (last lecture) yields

s !2 3 1 1010 ◦K t = + const = 1.78 sec + const 16πGag∗ T 2 T so   λ t ' 4.5 × 104 T X Ω h2 d 1010 p B which is substantially larger than unity until well after the temperature drops below 109 ◦K . Thus in this period the deuterium abundance is close to its equilibrium value √  Bd  Xd = 3 2XpXn ε exp kBT

9 ◦ Now the trouble: Td ' 0.7 × 10 K is low because Bd is small, hence deuteron

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 158 Cosmology remains rare until long after He4 would have been abundant in thermal equilibrium ⇒ processes 2) ⇒ stops nucleosynthesis (processes p + d → He3 + γ and n + d → H3 + γ have too low rates to help much)!

Ì But when T drops below Td the still present neutrons rapidly assembled in the most deeply bound light element He4! However, must overcome Coulomb barrier! Does not proceed via 2) but

p + n → Deuterium + n → Tritium + p → He4 or p + n → Deuterium + p → He3 + n → He4

Ì Now per proper volume: nHe helium, nH hydrogen nuclei: mass 4nHe + nH, hence fractional abundance by weight of helium Y = 4nHe/(4nHe + nH). Number of protons and neutrons np = 2nHe + nH and nn = 2nHe. Fraction of nucleons which are neutrons Xn = 2nHe/(4nHe + nH). Hence Y = 2Xn !

Ì formation of heavier nuclei blocked: no stable nuclei of atomic weight 5 and 8.

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 159 Cosmology

In stars formation of unstable Be8 in He4 + He4 collisions followed by resonant capture of another He4 to form excited state of C12

Ì When T ∼ Td the exothermic reactions 2) start to build up heavier elements: (cross sections ∝ 1/v). They are

hσ(d + d → H3 + p) vi ' 1.8 × 10−17 cm3/sec , hσ(d + d → He3 + n) vi ' 1.6 × 10−17 cm3/sec , and for the total rate one obtains

h 3 3 i λA3 = hσ(d + d → H + p) vi + hσ(d + d → He + n) vi Xd nN  T 3 = 1.9 × 107 sec−1 X Ω h2 1010 ◦K d B

This compares to the expansion rate after e+e−–annihilation:

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 160 Cosmology

 2 H = 1/2t = 0.28 T sec−1 1010 ◦K

9 ◦ for T in the region 10 K we have λA3 = H at

−7 2 −5 2 Xd ' 1.2 × 10 /ΩBh thus Xd ' 0.6 × 10 for ΩBh ' 0.02

This value is reached in thermal equilibrium at T about 109 ◦K and depends 2 weakly on ΩBh .

What it means: nucleosynthesis began around 109 ◦K and not at 0.75 × 109 ◦K . From table: t = 168 sec where Xn = 0.1609 × exp(−168/885) such that (p =primordial)

Yp ' 2 × 0.1609 × exp(−168/885) ' 0.27

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 161 Cosmology

Note:

« the larger the nucleon density the higher the temperature nucleosynthesis began

« the less time there was for neutrons to decay before nucleosynthesis

« the higher the He4 abundance

Results of modern state of the art calculations:

2 7 2 −10 ΩBh = 3.65 η × 10 [ΩBh = 0.02 ⇔ η = 5.5 × 10

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 162 Cosmology

Big Bang nucleosynthesis: reaction chains

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 163 Cosmology

Big Bang nucleosynthesis: predicted abundances

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 164 Cosmology

Data on He4/H ratio He4/H ratio: spectroscopy of so called HII regions (ionized gas containing low abundances of “metals” [elements other than hydrogen and helium], typically in blue compact galaxies) -≫

−10 Yp = 0.2477 ± 0.0029 corresponding to η = (5.813 ± 1.81) × 10

G The best Yp determination available, that by Peimbert et al. (2007a), is in agreement with the Dp determination and with the WMAP observations under the assumption of SBBN. The errors in the Yp determination are still large and there is room for non-standard physics. [“S”BBN=”Standard” Big Bang Nucleosynthesis]

G To improve the accuracy of the Yp value derived under the assumption of SBBN 2 and the ΩBh derived from the background radiation, a new determination of the neutron lifetime is needed to sort out the difference between the τn obtained by Arzumanov et al. (2000) and the τn obtained by Serebrov et al. (2005, 2008):

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 165 Cosmology

COSMOLOGICAL PREDICTIONS BASED ON SBBN AND OBSERVATIONS Table 1. FOR τn = 885.7 ± 0.8 s 2 Input Yp Dp η10 ΩBh a 2.53 b b b Yp 0.2477 ± 0.0029 2.93±1.06 5.625 ± 1.81 0.02054 ± 0.00661 b a b b Dp 0.2479 ± 0.0007 2.82 ± 0.28 5.764 ± 0.360 0.02104 ± 0.00132 b b b a WMAP 0.2487 ± 0.0006 2.49 ± 0.13 6.226 ± 0.170 0.02273 ± 0.00062 a b Observed value. Predicted value. References: τn Arzumanov et al. (2000); Yp Peimbert et al. (2007a); Dp O Meara et al.(2006); WMAP Dunkley et al. (2009).

Table 2. FOR τn = 878.5 ± 0.8 s 2 Input Yp Dp η10 ΩBh a 1.46 b b b Yp 0.2477 ± 0.0029 2.22±0.71 6.688 ± 1.81 0.02442 ± 0.00661 b a b b Dp 0.2462 ± 0.0007 2.82 ± 0.28 5.764 ± 0.360 0.02104 ± 0.00132 b b b a WMAP 0.2470 ± 0.0006 2.49 ± 0.13 6.226 ± 0.170 0.02273 ± 0.00062 a Observed value. b Predicted value. References: same as in Table 1 with the exception of τn that comes from Serebrov et al. (2005, 2008).

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 166 Cosmology

Characteristics of Yp: O The figure shows that Yp is not very sensitive to η, and therefor not a very good observable to constrain η.

O The helium rate is a good and important constraint on the expansion rate which yields T(t) in which g ∗ (T) plays a crucial role. One test here is the number of light neutrinos Nν. With Nν = 4 we would have g ∗ (T) = 3.813 instead of g ∗ (T) = 3.363, which would shorten the time as p t → t × g ∗ (Nnu = 3)/g ∗ (Nν = 4) = t × 0.94. This would lead to a higher helium abundance at the temperature 109 ◦K when nucleosynthesis begins.

Note: crucial interrelation between particle physics and cosmol- ogy/astrophysics: number of neutrinos, neutron lifetime!

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 167 Cosmology

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 168 Cosmology

Nν = 3.4

3.2

3.0

4 He production for Nν = 3.0, 3.2, 3.4. The vertical band indicates the baryon −5 density consistent with (D/H)p = (2.7 ± 0.6) × 10 and the horizontal line indicates a primeval 4He abundance of 25%. The widths of the curves in- dicate the 2 σ theoretical uncertainty [David N. Schramm and Michael S. Turner 1997].

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 169 Cosmology

Other light elements: Ì H2, H3, He3, Li7, and Be7 are produced in BBN. Ì The unstable nuclei H3 decayed later by β+–decay to He3, and Be7 by e−–capture to Li7. Ì The formation of He4 from free neutrons at T ∼ 109 ◦K of course was not 100% efficient, and free neutrons still were available to build other light nuclei via the two–body reactions of deuterium formation and deuterium fusion

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 170 Cosmology Data on D/H ratio The measurement of the deuterium abundance is complicated because of the small binding energy, which makes it easy to destroy it in stars. Modern techniques are based on observation of spectra from quasi-stellar objects.

D/H ratio: High-resolution spectra reveal the presence of D in high-redshift, low-metallicity quasar absorption systems via its isotope-shifted Lyman-α absorption. It is believed that there are no astrophysical sources of deuterium. Averaging the seven most precise observations of deuterium in quasar absorption systems gives

D/H = (2.82 ± 0.12) × 10−5 corresponding to η = (5.813 ± 1.81) × 10−10

-≫, -≫

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 171 Cosmology

Characteristics of D/H: O The figure shows that D/H is very sensitive to η, and therefore it is a very good observable to constrain η. The weak point is that its hard to be measured reliably. √  Bd  O The deuterium ratio is determined by Xd = 3 2XpXn ε exp (derived kBT above). The higher the baryon density, the more complete is the incorporation of neutrons into 4He, and hence the smaller the abundance of deuterium. As we −5 2 found above, at the start of BBN Xd ∼ 0.6 × 10 for ΩBh = 0.02. First Xd continued to rise for some time as the temperature dropped and the exponential term increased. But then Xd decreased again because Xn decreased due to the incorporation of free neutrons into deuterium and the conversion of deuterium to 3H and 3He, which convert to 4He. The final result of a detailed analysis is not 9 ◦ far from the deuterium fraction Xd we found at T = 10 K where we obtained −7 2 Xd ' 1.2 × 10 /ΩBh .

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 172 Cosmology

Data on He3/H ratio Data were obtained from galactic HII regions and from planetary nebulae. Problem: He3 is produced and destroyed in stars, and it is not clear whether the measured abundance is primordial. Quoted bound is

He3/H < (1.1 ± 0.2) × 10−5

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 173 Cosmology

Data on Li7/H ratio

2 7 Is interesting because of its more complicated dependence on ΩBh . Li is formed in two ways, by H3 + He4 → Li7 + γ and He3 + He4 → Be7 + γ, the latter followed much later by e− + Be7 → ν + Li7. At intermediate baryon densities Li7 is destroyed by p + Li7 → He4 + He4, which leads to the dip at η ∼ 2.5 × 10−10. Li7/H = (1.7 ± 0.06 ± 0.44) × 10−10

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 174 Cosmology Theory vs. Data: the BBN concordance range

ΩB ΩM Ω0

He4 (obs) ր

Deuterium (obs) ւ

He3 (bound) ր

Li7 (obs) ւ

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 175 Cosmology

Status Particle Data Group: (note different scales for different ratios)

-≫

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 176 Cosmology

Predicted abundances of 4He, D, 3He, and 7Li (bands show 95% CL range). Boxes show observed values (smaller boxes: ±2σ statistical errors; larger boxes: ±2σ statistical and systematic errors). Narrow vertical band: CMB measure of the cosmic baryon density; wider band: BBN concordance range (both at 95% CL).

For other summaries of BBN: -≫ -≫ -≫ -≫ or -≫

Conclusions:

O ∼ 25% 4He is a key piece of evidence for the Big Bang model, at ∼ 10% precision.

2 O D is the “baryometer”; best estimates of D/H imply ΩBh ≈ 0.02. O 3He and 7Li give two additional, low precision (factor ∼ 10) tests. Problem here: both elements can be produced and destroyed in stars.

BBN concordance provides a baryon content

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 177 Cosmology

2 0.019 ≤ ΩBh ≤ 0.024 at (95% CL)

The key result is the understanding of the baryonic matter budget (∼ 75% Hydrogen, ∼ 25% Helium plus well understood tiny contributions of other light elements, much less than 1% of all other elements combined). Note that ΩB  Ω0 far from being able to close the universe. Furthermore, the cosmic −1 density of (optically) luminous matter is Ωlum ' 0.0024 h , hence Ωlum  ΩB: most baryons are optically dark, probably in the form of a diffuse intergalactic medium, brown dwarfs, planets etc. Finally, given that ΩM ∼ 0.3, it appears that most matter in the Universe is not only dark, but also is non-baryonic.

ΩB ' 0.02 obtained from BBN is in good agreement with the independent determination from CMB [ΩB ' 0.0227 ]. As is well known ΩB is less than 10% of ΩM ' 0.3 [CMD gives ΩM ' 0.133 only], which includes any form of cold dark matter. The latter is still way below the critical density Ω0 = 1, at the limit of closing the universe and presumed by theory (inflation).

Actual cosmological parameters (PDG 2010):

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 178 Cosmology

Parameter Symbol Value Hubble parameter h 0.72 ± 0.03 2 Total matter density ΩM ΩMh = 0.133 ± 0.006 2 Baryon density ΩB ΩBh = 0.0227 ± 0.0006 Cosmological constant ΩΛ 0.74 ± 0.03 2 5 Radiation density Ωr Ωrh = 2.47 × 10 P 2 mν Neutrino density Ωνh = 93 eV Ων ∼ 0.001

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 179 Cosmology Summary and Outlook

As the universe expands, it cools. Free neutrons and protons are less stable than helium nuclei, and the protons and neutrons have a strong tendency to form helium−4. However, forming helium−4 requires the intermediate step of forming deuterium. At the time at which nucleosynthesis occurs, the temperature is high enough for the mean energy per particle to be greater than the binding energy of deuterium; therefore any deuterium that is formed is immediately destroyed (a situation known as the deuterium bottleneck). Hence, the formation of helium−4 is delayed until the universe becomes cool enough to form deuterium (at about T = 0.1 MeV), when there is a sudden burst of element formation. Shortly thereafter, at twenty minutes after the Big Bang, the universe becomes too cool for any nuclear fusion to occur. At this point, the elemental abundances are fixed, and only change as some of the radioactive products of BBN (such as tritium) decay.

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 180 Cosmology

Nucleosynthesis is predominantly He4 fusion

p p

p p n n p p n n n H2 H3 He4 n n

n n

n n p p n n p p p H2 He3 He4 p p

As the universe cools down for a short time protons and neutron can fuse to form light atomic nuclei

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 181 Cosmology

The Bottlenecks in BBN 8B 10B 11B 12B

7Be 8Be 9Be

6Li 7Li 8Li

3He 4He 5He

Elements of order ➄ and ➇ missing • 1H 2H 3H (totally unstable) Barrier for production • 1n of heavier elements in the early universe heavy elements must have been produced in stars • (extreme temperature and pressure) For element fusion in stars see “Neutrino Physics meets Astrophysics” in Lecture 3 -≫

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 182 Cosmology Primordial versus Stellar Nucleosynthesis Ì Timescale « Stellar Nucleosynthesis (SN): billions of years « Primordial Nucleosynthesis (PN): minutes 9 Ì Temperature evolution Be « SN: slow increase over time 7 « PN: rapid cooling Li 6 Ì Density Li « SN: 100 gr/cm3 −5 3 4He « PN: 10 gr/cm (like air) No stable 3 Ì Photon to baryon ratio He nuclei 2 « SN: less than a single photon per baryon H 1 « PN: billions of photons to every baryon H

The lack of stable elements of masses 5 and 8 makes it very difficult for BBN to progress beyond Lithium and even Helium. Need conditions found in stars for formation of the heavy elements.

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 183 Cosmology Nucleosynthesis by Neutron Capture

Stars like the sun use the proton-proton chain (main branch). Stars heavier than the sun (say 20 to 120 times the mass of the sun) use carbon-12 as a catalyst 12 (CNO-cycle). Very heavy stars use the triple-alpha process forming C6 from 3 Helium-4 nuclei (via intermediate Beryllium-8). Formation of elements beyond 56 iron ( Fe26) is possible only by neutron capture to produce isotopes.

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 184 Cosmology

Unstable isotopes decay into new elements. Example: 110 1 111 Cd48 + n0 → Cd48 neutron capture 111 1 112 Cd48 + n0 → Cd48 stable isotope 112 1 113 Cd48 + n0 → Cd48 stable isotope 113 1 114 Cd48 + n0 → Cd48 stable isotope 114 1 115 Cd48 + n0 → Cd48 unstable isotope 115 115 − Cd48 → In49 + e + ν¯e radioactive decay

Shell structure of a mass rich star:

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 185 Cosmology

Nucleosynthesis is rather well understood, the main problems seem to be the unambiguous determination of the element abundances (in particular 3He and Lithium). Nevertheless, results show a remarkable pattern for the abundances and the existence of a concordance range is highly non-trivial, as it covers almost 10 orders of magnitude in the element densities. Even more remarkable is the agreement with data obtained from the analysis of the CMB, which is completely independent. The CMB will be discussed below.

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 186 Cosmology Summary: “Cosmic Concordance” via BBN Ì Primordial nucleosynthesis « explains observed light element abundances if the density of normal matter (baryons) in the universe is in the range (3.2 − 4.5) × 10−31 gr/cm3 or 0.23 hydrogen atoms per cubic meter Ì Precise observational test « independent measurements of the abundances of the four lightest stable elements lead to consistent constraints on the density of normal matter « provides confidence that primordial or Big Bang nucleosynthesis provides the correct explanation of the formation of the light elements

With BBN we have quited the pure particle physics era and entered the domain of theoretically more involved nuclear physics. At this stage only the simplest nuclei come into play, this allows us to make rather unambiguous predictions. The trouble with clear cut experimental tests follows where elements produced in primordial nucleosynthesis later also were produced in stars which makes the interpretation of observed abundances difficult.

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 187 Cosmology Exercise pages

Exercise: check (numerically) that for fixed time–independent temperature T = Tν λ(p→n) the two rates λ(p → n) and λ(n → p) have a ratio λ(n→p) = exp (−Q/kBT) for T = Tν.

Exercise: show that Xn → 0.1609 exp (−t/τn), with τn = 885.7 sec is the neutron lifetime, fits the tabulated values of Xn for not too small times.

Exercise: at some time in the expansion of the universe photons were in thermal equilibrium with non-relativistic electrons (matter in general). Photons then followed the Planck black-body radiation equilibrium distribution with equilibrium temperature Tγ = Te. As expansion continues, assume (for simplicity) that at some fixed time tL (time of last-scattering) photon stop to interact with matter, because photon energy falls below the interaction threshold energy. Show that, up to a change in temperature, the form of the distribution remains unchanged by the expansion of the universe even though photons get out of equilibrium with matter. How does the temperature change?

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 188 Cosmology

Previous ≪R , next R≫ lecture.

c 2009, F. Jegerlehner ≪R Lect. 9 R≫ 189